Then I subtract line 2 from line one, and line 2 from line 3 which leaves:
2 1 0 -2
0 3 6 -6
0 0 0 0

If this were in fact correct, you could divide the second line by 3 to get
2 1 0 -2
0 1 2 -2
0 0 0 0
then subtract the second line from the first to get
2 0 -2 0
0 1 2 -2
0 0 0 0

which corresponds to the equations 2x- 2z= 0 and y+ 2z= -2. That would mean that you can write both x and y in terms of z: x= z and y= -2- 2z. If take z itself to be a parameter, that's the equation of a line: x= t, y= -2- 2t, z= t. Here the three planes intersect in a single line.

But, as Prove It said, you have made an arithmetic error. In fact, these three planes do intersect in a single point.

If this were in fact correct, you could divide the second line by 3 to get
2 1 0 -2
0 1 2 -2
0 0 0 0
then subtract the second line from the first to get
2 0 -2 0
0 1 2 -2
0 0 0 0

which corresponds to the equations 2x- 2z= 0 and y+ 2z= -2. That would mean that you can write both x and y in terms of z: x= z and y= -2- 2z. If take z itself to be a parameter, that's the equation of a line: x= t, y= -2- 2t, z= t. Here the three planes intersect in a single line.

But, as Prove It said, you have made an arithmetic error. In fact, these three planes do intersect in a single point.

This is exactly what I wanted to hear! I was given diagrams of graphs to match this to, and the one I chose was the one that had the 3 planes intersecting in a line.

Thank you all for your help!!! Sorry for the typo in my original post, proves you are all very keen in your arithmetic!